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$\DeclareMathOperator*{\plim}{plim}$ I thought I understood the idea of convergence in probability. I know the definition is $$\plim_{n\to\infty} X_n\rightarrow X \;\; \text{if and only if} \;\; \lim_{n\to\infty}\mathrm{Pr}(|X_n-X|\geq\epsilon)=0 \, ,\;\; \text{for every} \;\; \epsilon>0\, .$$ I kind of understand the definition and what it says, but I am unsure how to get it. For example, I get that $$\plim_{n\to\infty} \bar{X}_n=\mu\, .$$ But how would I calculate something such as $$\plim_{n\to\infty} \frac{1}{n}\sum_{i=1}^n X_i^2= \mathrm{?}$$

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    $\begingroup$ If you write $y_i = x_i^2$ then the expression you ask about is $\bar y$ and so, apart from notation, it doesn't seem to be anything different from the previous limit which you say you understand. That makes it difficult to gauge your question: perhaps you have already answered it; perhaps it asks about how to compute limits of sequence of random variables quite generally (and therefore would be far too broad and difficult to address adequately in a forum like this). Could you edit this question to make it more specific? $\endgroup$ – whuber Oct 3 '14 at 15:45
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    $\begingroup$ So would it just converge to E($X^2$) = Var(X)+$\mu_x^2$? $\endgroup$ – yankeefan11 Oct 4 '14 at 16:36
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Yes, what you say in your last comment is basically true. You didn't get any answers for so long because your post does not include enough context/assumptions. But you seems to assume that the variables $X_1, X_2, \dotsc$ all are independent and with the same distribution (IID) as $X$, and then you just apply the (strong) law of large numbers and can conclude, as you did in comment, that $$ \plim_{n\to\infty} \frac{1}{n}\sum_{i=1}^n X_i^2 = \mathbb{E} X^2= \sigma^2_x+\mu_x^2 $$

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